Question

39. An electric iron consumes energy at a rate of 800 W when heating at the maximum rate and 300 W
when heating is at the minimum rate. The voltage is 200 V. What are the current and resistance in each
case?

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Answer 1

case1

P=800w

V=200v

P=V×I

I=P/V= 800/200= 4Amp

P=I²×R

R=P/I²=800/4²=100/2=50ohms

case2

P=300w

V=200v

P=V×I

I=P/V= 300/200= 1.5Amp

R=P/I²= 300/ 9/4= 133.333ohms

Answer 2

Given,

Energy consumed by iron at rate when heat is at maximum rate = 800W

Energy consumed by iron at rate when heat is at minimum rate = 300W

Voltage = 200V

To find,

Resistance and current in both cases

Solution,

Case 1

P = v x l

l = p/v

= 800/200

= 4Amp

P = I² x R

R = P/I²

= 800/4²

= 100/2

= 50ohm

Case 2

I = P/V

= 300/200

= 1.5Amp

R = P/I²

= 300/(9/4)

= 133.33 Ohms

Thus, the resistance in the first case is 50ohm and current is 4A. In the 2nd case the current is 1.5A and resistance is 133.33Ohm.